AC Output Voltage as function of frequency

Click For Summary

Homework Help Overview

The discussion revolves around deriving an expression for the output voltage of a sine-wave generator connected to a high-pass filter, specifically focusing on how the output voltage varies with frequency. The original poster expresses confusion regarding the wording of the problem and the transition between time and frequency domains.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss applying Kirchhoff's laws and Ohm's law to derive a differential equation relating the input and output voltages. There are attempts to manipulate equations involving trigonometric functions and concerns about transitioning from time domain to frequency domain representations.

Discussion Status

Participants are actively exploring different methods to approach the problem, with some suggesting the use of phasors to simplify calculations. There is recognition of the need to express the output voltage in terms of frequency, and several participants are questioning their assumptions and the correctness of their equations.

Contextual Notes

There are indications of confusion regarding the treatment of the capacitor's voltage and the initial conditions for integration. Participants are also grappling with sign errors in their equations and the implications of the circuit elements involved.

kenef
Messages
10
Reaction score
0

Homework Statement


Given a sine-wave generator of frequency f having internal resistance R1 connected to a high-pass filter. If the generator shown in the dotted outlined box produces a voltage V0(t) = V0 cos(2π ft) with no load, derive an expression for the output voltage V1(t) = V1 cos(2π ft + φ) as a function of frequency f.

Homework Equations


V= I R
Xcap = 1 / ωC

The Attempt at a Solution


[/B]
Ultimately I am lost at the wording in the question. If one is supposed to take the V1 as the accurate final voltage and simply mathematically manipulate it that is what I have attempted. I know that cos(A+B) = (cos(A)cos(B) - sin(A)cos(B) however from there I do not see much I am allowed to do. Perhaps it is my math that is a little rusty. I do remember a long time ago performing manipulations of expressing one variable in terms of another however involving trigonometric functions seems a bit confusing to me.

I know another method would be to go from this time domain to the frequency domain however this manipulation is unknown to me, as the book we are using only talks about the basis of Time Domain view of RCs and Frequency domain view of RCs... " Learning the Art of Electronics - Hayes"
 

Attachments

  • IMG_2394.jpg
    IMG_2394.jpg
    81.5 KB · Views: 567
Last edited:
Physics news on Phys.org
You want to apply Kirchoff's voltage and current laws and Ohm's law to derive a differential equation that relates V0 and V1.

Alternately, you can avoid some of the math if you're familiar with phasors.

https://en.wikipedia.org/wiki/Phasor
 
vela said:
You want to apply Kirchoff's voltage and current laws and Ohm's law to derive a differential equation that relates V0 and V1.

Alternately, you can avoid some of the math if you're familiar with phasors.

https://en.wikipedia.org/wiki/Phasor
I was going to do this approach, however once I relate V0 and V1, I am still left with equations in which are of the time domain and not frequency domain
 
Why is that a problem?
 
Because I need the output voltage in terms of the frequency and not time as it states, at least that's what it means to me. Doing this I am left with the a function in time not frequency...

Maybe I am just over thinking this as well
 
The coefficient of the cosine in V1(t) will depend on frequency. That's what the problem means by as a function of frequency. You can still solve the problem in the time domain.
 
  • Like
Likes   Reactions: kenef
Ok so far, I have that V0 cos(2π ft) + R1C dV/dt = V1 cos(2π ft + φ) . Is it correct to assume that R2 should not be considered ?

With this in mind I do know that my final relation should show me what V1/V0 is equal to.

Update:

I have that V0 cos(2π ft) + (R1 + R2) C dV/dt = V1 cos(2π ft + φ), with manipulation I have arrived to V1cosφ - V0 = C( R1 + R2) ∫dV(t) . I do not feel too confident with my approach in this manner, because I have no initial and final values in order to integrate
 
Last edited:
Your differential equation isn't quite right. It's probably simpler if you write a differential equation for the voltage across the capacitor (what you called V) and then relate it to V1. The current in the circuit is ##i = C \frac{dv}{dt}## where ##v## is the current across the capacitor. Applying Ohm's Law to R2, you get ##V_1 = iR_2 = R_2 C\frac{dv}{dt}##. So once you find ##v##, you'll essentially know ##V_1##.

To get the differential equation that governs ##v##, you want to apply KVL to the circuit. You kind of did it, but you made errors. For one thing, source elements and sink elements (like resistors and capacitors) should have different signs. You're also missing the voltage drop across the capacitor.
 
You are right, I have not done any circuits in a while and completely missed the fact that I have to subtract my voltage drops across the sink elements! As far as the voltage drop across the capacitor I attempted to perform something that I guess is incorrect, I thought of the capacitor as not charged. Ok so far I have the V0 cos(2π ft) - C {dv}/{dt} (R_1 - R_2 - 1) = 0, is this along the correct path? Doing this I ended up with a rather weird derivation.

I attempted it in the method you described, using only V1 and R2, I came to a solution of V1 = e^ ( (t+ const) / R_2 * C))
 
Last edited:
  • #10
The plates of the capacitor may not be initially charged, but they acquire a charge as current flows in one end and out the other.

You still have some sign errors, and the constant 1 you threw in there isn't correct. (It doesn't work unit-wise for one thing.) The voltage across the capacitor is simply v; it's not C dv/dt. Keep trying. You're really close.
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
3K
Replies
17
Views
3K
Replies
2
Views
2K
  • · Replies 47 ·
2
Replies
47
Views
4K
Replies
18
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
10K
Replies
1
Views
2K